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Aflati numerele naturale x,y,z stiind ca sunt respectiv proportionale cu 0,25; 0,(3); 0,5 si ca xy +xz +yz =54.

Răspuns :

Mariangelask
0,25=1/4
0,(3)=1/3
0,5=1/2
"x,y,z stiind ca sunt respectiv proportionale cu 0,25; 0,(3); 0,5" inseamna ca" exista k nr nat astfel incat:
x/0,25=y/0,(3)=z/0,5=k, adica
x/(1/4)=y/(1/3)=z/(1/2)=k si facand calculele:
4x=3y=2z=k, deci:
x=k/4, y=k/3 si z=k/2 si inlocuim in relatia:

xy +xz +yz =54
k*k/(3*4)+k*k/(2*4)+k*k/(3*2)=54 Aducem la acelasi numitor:
k*k(2+3+4)/(2*3*4)=9*6
k*k*9/(2*3*4)=9*6 impartim prin 9 si unmultim cu 2*3*4:
k*k=6*6*4=144 deci k=12, de unde x=3, y=4, z=6








Crush19ask
0,25=[tex] \frac{25}{100} = \frac{1}{4} [/tex]

[tex]0,(3) = \frac{3}{9} = \frac{1}{3} [/tex]

[tex] 0,5= \frac{5}{10} = \frac{1}{2} [/tex]

[tex] \frac{x}{ \frac{1}{4} } = \frac{y}{ \frac{1}{3}} = \frac{z}{ \frac{1}{2} } [/tex] = k

[tex] \frac{x}{1} : \frac{1}{4} = \frac{y}{1} : \frac{1}{3} = \frac{z}{1} : \frac{1}{2} = k[/tex]

4x = 3y = 2z = k

[tex]x = \frac{k}{4} [/tex]

[tex]y = \frac{k}{3} [/tex][tex] \frac{k}{4} [/tex]

[tex]z = \frac{k}{2} [/tex]

[tex] \frac{k}{4} [/tex] * [tex] \frac{k}{3} [/tex] + [tex] \frac{k}{4} [/tex] * [tex] \frac{k}{2}
[/tex] + [tex] \frac{k}{3} [/tex] * [tex] \frac{k}{2} [/tex] = 54

[tex] \frac{k^{2} }{12} + \frac{k^{2} }{8} + \frac{k^{2} }{6} = 54[/tex]

aducem la acelasi numitor

[tex] \frac{2*k^{2} }{24} + \frac{3*k^{2} }{24} + \frac{4*k^{2} }{24} = 54[/tex]

[tex] \frac{9*k^{2}}{24} = 54 [/tex]

[tex]9* k^{2} = 24*54 [/tex]

[tex]9* k^{2} = 1296 [/tex]        /:9

[tex]k^{2} =144[/tex]

k=12

x=12:4=3

y=12:3=4

z=12:2=6







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